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When going through one of my chemistry textbooks, I saw that the electrolysis of aqueous lead nitrate led to oxygen being formed at the anode and lead being formed at the cathode.

However, in school, I was taught that only metals less reactive than hydrogen will form at the cathode. This electrolysis contradicts what I learned, as lead is more reactive than hydrogen, and should remain as ions in solution.

Therefore, my question is: What forms at the cathode during electrolysis of aqueous lead nitrate and why?

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Firstly, the reactivity being referred to is not reactivity in the sense of the reactivity series of metals, but rather in the sense of standard electrode potentials (standard reduction potentials).

This comes into play when setting up an electrolytic cell in determining the reaction that will take place at the cathode/anode. In this case, the reaction that requires the least amount of work (or voltage in this case, since voltage is essentially a measure of work) will take place. So, assuming that this is done aqueously, if the reduction/oxidation reaction to take place requires more voltage than the reduction/oxidation of the water, then the water will simply be reduced/oxidized instead. (I think this is where the reactivity of hydrogen comes into play in how you were taught).

So to answer your question more directly, it is actually lead that forms at the cathode.

Edit: To add some detail, the reduction potential of H2O is -0.8277 V, while the lead(II) ion is -0.13. In this case, less energy means closer to zero, the negative being essentially irrelevant. Hope this helped.

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